Home

Let $X$ be a Hilbert space and $T ∈ ℬ(X)$. Show that the graph $Γ(T)$ of $T$ is a closed subspace of $X × X$ and that \[ Γ(T)^⟂=\{(-T^* x, x): x ∈ X\} . \] By considering the orthogonal decomposition of $(x, 0)$, prove that $I+T^* T: X → X$ is surjective. [Recall: if $X$ is Hilbert space, $X × X$ with $⟨(x_1, x_2),(y_1, y_2)⟩_{X × X}=⟨x_1, y_1⟩_X+⟨x_2, y_2⟩_X$ is Hilbert space.] Proof. $Γ(T)$ is closed by Closed Graph Theorem. $$(a,b)∈Γ(T)^⟂⇔∀x,⟨(a,b),(x,T(x))⟩=⟨a,x⟩+⟨b,T(x)⟩=0\\⇔∀x,⟨a+T^*(b),x⟩=0\\⇔a=-T^*(b).$$ so $Γ(T)^⟂=\{(-T^* x, x): x ∈ X\}.$ Let the orthogonal decomposition of $(x, 0)$ be $(x, 0)=(y,Ty)+(x-y,-Ty)$. $(x-y,-Ty)∈Γ(T)^⟂$, so $x-y=T^*Ty$, so $x∈\operatorname{Im}(I+T^* T)$, so $I+T^* T$ is surjective.

Let $P(x, y, z)$ be a homogeneous polynomial of degree $d$ defining a nonsingular curve $C$ in $ℂℙ^2$. (i) Write down Euler's relation for $P, P_x, P_y, P_z$. Deduce that the Hessian determinant satisfies: \[ z ℋ_P(x, y, z)=(d-1) \det\begin{pmatrix} P_{xx} & P_{xy} & P_{xz} \\ P_{yx} & P_{yy} & P_{yz} \\ P_x& P_y & P_z \end{pmatrix} \] Solution. Euler's relation $dP=xP_x+yP_y+zP_z$. Both sides $∂_x$ we get $dP_x=P_x+xP_{xx}+yP_{yx}+zP_{zx}$, so \[(d-1)P_x=xP_{xx}+yP_{yx}+zP_{zx}\] Both sides $∂_y$ we get $dP_y=P_y+xP_{xy}+yP_{yy}+zP_{zy}$, so \[(d-1)P_y=xP_{xy}+yP_{yy}+zP_{zy}\] Both sides $∂_z$ we get $dP_z=P_z+xP_{xz}+yP_{yz}+zP_{zz}$, so \[(d-1)P_z=xP_{xz}+yP_{yz}+zP_{zz}\] By definition of Hessian and linearity of det, \[ z ℋ_P(x, y, z)=\det\begin{pmatrix} P_{xx} & P_{xy} & P_{xz} \\ P_{yx} & P_{yy} & P_{yz} \\ zP_{zx} &zP_{zy}&zP_{zz} \end{pmatrix}\] Adding $x$ times first row, $y$ times second row to third row \[=(d-1) \det\begin{pmatrix} P_{xx} & P_{xy} & P_{xz} \\ P_{yx} & P_{yy} & P_{yz} \\ P_x& P_y & P_z \end{pmatrix}. ∎ \] (ii) Deduce further that: \[ z^2 ℋ_P(x, y, z)=(d-1)^2 \det\begin{pmatrix} P_{xx} & P_{xy} & P_x\\ P_{yx} & P_{yy} & P_y \\ P_x& P_y & d P /(d-1) \end{pmatrix} \] Solution. Adding $x$ times first column, $y$ times second column to third column, we get the determinant. ∎ (iii) Deduce that if $P(x, y, 1)=y-g(x)$ then $[a, b, 1]$ is a point of inflection of $C$ if and only if $b=g(a)$ and $g''(a)=0$. This shows the lectures definition of points of inflection corresponds to the usual notion of a point of inflection of the graph of a function $g(x)$ on $ℝ$ or $ℂ$. Solution. point$∈C⇔P(a,b,1)=0⇔b=g(a)$. By (ii), \[ z^2 ℋ_P(x, y, z)=(d-1)^2 \det\begin{pmatrix} -g''(x)&0&-g'(x)\\ 0&0&1\\ -g'(x)&1&0 \end{pmatrix}=(d-1)^2g''(x) \] so $ℋ_P(a,b,1)=0⇔g''(a)=0$ (if $d=1$, $g''=0$).

Let $X$ and $Y$ be real Hilbert spaces and let $T ∈ ℬ(X, Y)$ be surjective. Show that there exists a unique bounded linear operator $R ∈ ℬ(Y, X)$ such that \[ T R=I_Y \text{ and }\|R T x\|≤\|x\| \text{ for all } x ∈ X . \] %[Hint: Follow the proof 3.11 that for operators between Hilbert spaces $S X$ is closed iff $S^* Y$ is closed] Proof. Define $Z=\ker(T)^⟂⊆X,A=T|_Z∈ℬ(Z,Y)$. $A$ is injective since $Z∩\ker(T)=\{0\}$. $A$ is surjective since $Tx=T(P^Zx)=A(P^Zx)∀x∈X$ where $P^Z$ is orthogonal projection onto $Z$. $A$ is bijective, hence $∃A^{-1}∈ℬ(Y,Z)$ by Inverse Mapping Theorem. Define $R∈ℬ(Y,X),Ry=A^{-1}y∀y∈Y$, then $TR=AA^{-1}=I_Y$. $RT(x)=RA(P^Zx)=P^Zx$, so $‖RT(x)‖≤‖x‖∀x∈X$. Uniqueness: Say $\tilde{R}$ is another possible candidate. For all $y∈Y$, \begin{align*}‖P^Z\tilde{R}y‖&≥‖\tilde{R}TP^Z\tilde{R}y‖ &\text{since }&\|x\|≥\|\tilde{R}T x\|\text{ for all } x ∈ X \\&=‖\tilde{R}T\tilde{R}y‖&\text{since }&TP^Z=T \\&=‖\tilde{R}y‖&\text{since }&T\tilde{R}=I_Y&\end{align*} but $‖\tilde{R}y‖≥‖P^Z\tilde{R}y‖$ by Pythagoras, so $\tilde{R}y∈Z$, so $$ARy=y=T\tilde{R}y=A\tilde{R}y$$ but $A$ is bijective, so $Ry=\tilde{R}y$.

Fourier Series, Fourier Transform and Their Applications to Mathematical Physics pp 33–35 Theorem 6.1 (Riemann–Lebesgue lemma). If f is periodic with period $2π$ and belongs to $L^1(-π ,π )$, then \begin{align*}\tag{6.1}\lim _{n→ ∞ }∫_{-π }^π f(x+z)\mathrm{e}^{-\mathrm{i}nz}\mathrm{d}z=0\label{Equ1}\end{align*} uniformly in $x∈ ℝ$. In particular, $c_n(f)→ 0$ as $n→ ∞$. Proof. Since f is periodic with period $2π$, it follows that \begin{align*}\tag{6.2}\label{Equ2}∫_{-π }^π f(x+z)\mathrm{e}^{-\mathrm{i}nz}\mathrm{d}z=∫_{-π +x}^{π +x} f(y)\mathrm{e}^{-\mathrm{i}n(y-x)}\mathrm{d}y=\mathrm{e}^{\mathrm{i}nx}∫_{-π }^{π } f(y)\mathrm{e}^{-\mathrm{i}ny}\mathrm{d}y \end{align*} by Lemma 1.3. Formula ($\ref{Equ2}$) shows that to prove ($\ref{Equ1}$) it is enough to show that the Fourier coefficients $c_n(f)$ tend to zero as $n→ ∞$. Indeed, \[2π c_n(f)=∫_{-π }^π f(y)\mathrm{e}^{-\mathrm{i}ny}\mathrm{d}y=∫_{-π +π /n}^{π +π /n} f(y)\mathrm{e}^{-\mathrm{i}ny}\mathrm{d}y=∫_{-π }^π f(t+π /n)\mathrm{e}^{-\mathrm{i}nt}\mathrm{e}^{\mathrm{i}π }\mathrm{d}t\] by Lemma 1.3. Hence \begin{align*}\label{Equ3}\tag{6.3} -4π c_n(f)=∫_{-π }^π \left( f(t+π /n)-f(t)\right) \mathrm{e}^{-\mathrm{i}nt}\mathrm{d}t. \end{align*} If f is continuous on the interval $[-π ,π ]$, then \[\sup _{t∈ [-π ,π ]}\left| f(t+π /n)-f(t)\right| → 0,  n→ ∞ .\] Hence $c_n(f)→ 0$ as $n→ ∞$. If f is an arbitrary $L^1$ function, we let $ε >0$. Then we can define a continuous function g (see Corollary 5.3) such that \[∫_{-π }^π \left| f(x)-g(x)\right| \mathrm{d}x<ε .\] Write \[c_n(f)=c_n(g)+c_n(f-g).\] The first term tends to zero as $n→ ∞$, since g is continuous, whereas the second term is less than $ε /(2π )$. This implies that \[\lim _{n→ ∞ }|c_n(f)|=0.\] This fact together with ($\ref{Equ2}$) gives ($\ref{Equ1}$). The theorem is thus proved. $□$ Corollary 6.2. Let f be as in Theorem $\ref{Equ1}$. If a periodic function g is continuous on $[-π ,π ]$, then \[\lim _{n→ ∞ }∫_{-π }^π f(x+z)g(z)\mathrm{e}^{-\mathrm{i}nz}\mathrm{d}z=0\] and \begin{align*} \lim _{n→ ∞ }∫_{-π }^π f(x+z)g(z)\sin (nz)\mathrm{d}z=\lim _{n→ ∞ }∫_{-π }^π f(x+z)g(z)\cos (nz)\mathrm{d}z=0 \end{align*} uniformly in $x∈ [-π ,π ]$. Exercise 6.1. Prove this corollary. Exercise 6.2. Show that if f satisfies the Hölder condition with exponent $α ∈ (0,1]$, then $c_n(f)=O(|n|^{-α })$ as $n→ ∞$. Exercise 6.3. Suppose that f satisfies the Hölder condition with exponent $α >1$. Prove that $f≡ \text {constant}$. Exercise 6.4. Let $f(x)=|x|^α$, where $-π ≤ x≤ π$ and $0<α <1$. Prove that $c_n(f)≍ |n|^{-1-α }$ as $n→ ∞$. Remark 6.3. The notation $a≍ b$ means that there exist $0modulus of continuity of f by \begin{align*} ω _{p,δ }(f):=\sup _{|h|≤ δ } \left( ∫_{-π }^π \left| f(x+h)-f(x)\right| ^p\mathrm{d}x\right) ^{1/p}. \end{align*} The equality ($\ref{Equ3}$) leads to \begin{align*} |c_n(f)|&≤ \frac{1}{4π }∫_{-π }^π \left| f(x+π /n)-f(x)\right| \mathrm{d}x\\&≤ \frac{(2π )^{1-1/p}}{4π }\left( ∫_{-π }^π \left| f(x+π /n)-f(x)\right| ^p\mathrm{d}x\right) ^{1/p}≤ \frac{1}{2}(2π )^{-1/p}ω _{p,π /n}(f), \end{align*} where we have used Hölder’s inequality in the penultimate step. Exercise 6.5. Suppose that $ω _{p,δ }(f)≤ Cδ ^α$ for some $C>0$ and $α >1$. Prove that f is constant almost everywhere. Hint. First show that $ω _{p, 2δ }(f)≤ 2ω _{p,δ }(f)$; then iterate this to obtain a contradiction. Suppose that $f∈ L^1(-π ,π )$ but f is not necessarily periodic. We can consider the Fourier series corresponding to f, i.e., \[f(x)∼ \sum _{n=-∞ }^∞ c_n \mathrm{e}^{\mathrm{i}nx},\]where the $c_n$ are the Fourier coefficients $c_n(f)$. The series on the right-hand side is considered formally in the sense that we know nothing about its convergence. However, the limit \begin{align*}\label{Equ4}\tag{6.4}\lim _{N→ ∞ }∫_{-π }^π \sum _{|n|≤ N} c_n(f) \mathrm{e}^{\mathrm{i}nx}\mathrm{d}x=∫_{-π }^π f(x)\mathrm{d}x \end{align*} exists. Indeed, \begin{align*} ∫_{-π }^π \sum _{|n|≤ N} c_n(f) \mathrm{e}^{\mathrm{i}nx}\mathrm{d}x&=c_0(f)∫_{-π }^π \mathrm{d}x +\sum _{0<|n|≤ N}c_n(f)∫_{-π }^π \mathrm{e}^{\mathrm{i}nx}\mathrm{d}x=2π c_0(f)\\&=∫_{-π }^π f(x)\mathrm{d}x. \end{align*} Remark 6.4 (Important properties of the Fourier series). The existence of the limit ($\ref{Equ4}$) shows us that we can always integrate the Fourier series of an $L^1$ function term by term.

Proposition 3.15. A nonsingular points $p$ in a curve $C$ in $ℂ ℙ^2$ is a point of inflection if $I_p(C, L) ⩾ 3$, where $L$ is the tangent line to $C$ at $p$. Proof. After a projective transformation can take $P=[0,0,1]$ and $L=\{x=0\}$. Then \[ P(0,0,1)=P_y(0,0,1)=P_z(0,0,1)=0 \text {, and } P_x(0,0,1) ≠ 0 \] as $P$ is nonsingular. So by Lemma 3.14 \[ H_p(0,0,1)=(d-1)^2 \det\left(\begin{array}{ccc} P_{x x}(0,0,1) & P_{x y}(0,0,1) & P_x(0,0,1) \\ P_{y x}(0,0,1) & P_{y y}(0,0,1) & 0 \\ P_x(0,0,1) & 0 & 0 \end{array}\right) =-(d-1)^2P_x(0,0,1)^2 P_{y y}(0,0,1) . \] Thus $p$ is a point of inflection if $P_{y y}(0,0,1)=0$. But $I_P(C, L)$ is the largest power of $(y-0 z)$ dividing $R_{P(x, y, z), x}=P(0, y, z)$ As $P(0,0,1)=P_y(0,0,1)=0$, $y^2$ divide $P(0, y, z)$, and $y^3$ divides $P(0, y, z)$ if $P_{yy}(0,0,1)=0$. Here $I_p(C, L) ⩾ 2$, and $I_p(C, L) ⩾ 3$ if $p$ is a point of inflection. ∎

Derive the open mapping theorem from the inverse mapping theorem in the case $X$ is a Hilbert space. Proof. $X$ is a Hilbert space, $Y$ is a Banach space, $T\colon X→Y$ is a surjective linear map. By Theorem 3.11, $TX$ closed implies $T^*Y$ closed. Let $Q\colon X→T^*Y$ be the orthogonal projection. $Q$ is open: $B^{T^*Y}=Q(B^{T^*Y})⊆Q(B^X)$. Let $S≔T|_{T^*Y}$. $S\colon T^*Y→Y$ is injective: if $x∈\ker(S)$ then $x∈\ker(T)∩T^*Y=\{0\}$. By the inverse mapping theorem $S$ has a continuous inverse, so $S$ is open. Since $\ker(T)^⟂=T^*Y$, for all $x∈X,T(x-Qx)=0⇒Tx=TQx$, so $T=S∘Q$ is open. % https://math.stackexchange.com/questions/4330870/inverse-mapping-theorem-implies-open-mapping-theorem [In general: Quotient map is surjective so open. Projection map is surjective so open.]

Proof: By the primality of $p$ and the Sylow theorems, we have that the $p$-Sylow subgroups are all generated by elements of order $p$ that are conjugate to each other. Therefore, we can take without loss of generality our element of order $p$ to be the cycle $(234\dots p1)$. Let our transposition be $(ij)$. We can conjugate by our cycle to generate any transposition of $k$ and $k + j - i$. Then we can compose these transpositions of size $j - i$ starting at $1$, using primality at $p$ to get a transposition between $1$ and $h$ for $h < j - i$. This continues until we get the transposition $(12)$. Conjugating by our cycle repeatedly gives us all of the transpositions between $i$ and $i+1$, which then generate $S_p$.

Proposition 16 A nonsingular point $p \in C$ is an inflection point if and only if $\det P_{i j}=0$. Proof: To find the multiplicity of $p$ with respect to a line we have to take the resultant of $P(x, y, z)$ and $a x+b y+c z$. But this is just substituting $x=-(b y+c z) / a$ into $P$. It is more convenient to retain the symmetry and consider the line as the set of points \[ \left[a_0+t \alpha_0, a_1+t \alpha_1, a_2+t \alpha_2\right] \] as $t$ varies. Then $I_p(C, L) \geq 3$ if and only if $t^3$ divides \[ P\left(a_0+t \alpha_0, a_1+t \alpha_1, a_2+t \alpha_2\right) . \] Expanding this, we have \[\tag7 P(a+t \alpha)=P(a)+t \sum_i P_i(a) \alpha_i+\frac{t^2}{2} \sum_{i, j} P_{i j}(a) \alpha_i \alpha_j+t^3 R \] and $t^3$ divides this if and only if \[ \sum_i P_i(a) \alpha_i=0=\sum_{i, j} P_{i j}(a) \alpha_i \alpha_j \] The first equation says that the line is a tangent. We now use homogeneity - the $P_i$ are homogeneous of degree $n-1$ - so Euler's relation gives \[ (n-1) P_i(a)=\sum_j P_{i j}(a) a_j \] This always gives \[\tag8 \sum_{i, j} P_{i j}(a) a_i a_j=(n-1) \sum_i P_i(a) a_i=n(n-1) P(a)=0 \] and in our case also \[ \sum_{i, j} P_{i j}(a) a_i \alpha_j=(n-1) \sum_i P_i(a) \alpha_i=0 \] from the first equation. Together with the second equation we see that the quadratic form on the vector space spanned by $a$ and $\alpha$ (the subspace of $\mathbf{C}^3$ defining the line $L$) vanishes completely. This means that the matrix of the quadratic form with respect to a basis $a, \alpha, \beta$ is of the form \[ \left(\begin{array}{lll} 0 & 0 & * \\ 0 & 0 & * \\ * & * & * \end{array}\right) \] and so $\det P_{i j}(a)=0$. [A small lemma: a bilinear form vanish on a line then it vanish.] Conversely take a basis of $a, \alpha, \beta$ where $a$ and $\alpha$ define the tangent at $p$. From (8) the matrix is of the form \[ \left(\begin{array}{lll} 0 & 0 & * \\ 0 & * & * \\ * & * & * \end{array}\right) \] If $\sum_{i j} P_{i j} a_i \beta_j=0$ then $\sum_i P_i \beta_i=0$ by homogeneity, but then $\left[\beta_0, \beta_1, \beta_2\right]$ lies on the tangent so $a, \alpha, \beta$ do not form a basis. Hence the determinant $\det P_{i j}$ vanishes if and only if the central term \[ \sum_{i, j} P_{i j}(a) \alpha_i \alpha_j=0 \]

Let $X$ and $Y$ be real Banach spaces and $T ∈ ℬ(X, Y)$. Assume that $Z=T X$ is a finite-codimensional subspace of $Y$ and let $\{y_1+Z, …, y_m+Z\}$ be a basis for $Y / Z$. Define $\hat{T}: X ⊕ ℝ^m → Y$ by \[ \hat{T}(x,(v_1, …, v_m))=T(x)+\sum_{j=1}^m v_j y_j . \] Show that $\hat{T}$ is a surjective bounded linear operator. Hence, by applying the open mapping theorem, deduce that $Z$ is closed. Proof: Clearly $\hat{T}$ is linear. \[ ‖\hat{T}(x,(v_1, …, v_m))‖≤‖T(x)‖+‖\sum_{j=1}^m v_j y_j‖ ≤‖T‖‖x‖+\max_j‖y_j‖\sum_{j=1}^m‖v_j‖ \] so $\hat{T}$ is bounded. For any $y∈Y$, let $y+Z=\sum_{j=1}^m v_j(y_j+Z)$, then $y-\sum_{j=1}^m v_jy_j∈Z=TX$, so $y-\sum_{j=1}^m v_jy_j=T(x)$ for some $x$, so $y=\hat{T}(x,(v_1, …, v_m))$, so $T$ is surjective. By open mapping theorem, $\hat{T}$ is open, so $\hat{T}(X,ℝ^m∖\{0\})=Y∖Z$ is open, so $Z$ is closed.